Question

ABC is a triangle in which AB=AC and P is any point on AC. Through C a line is drawn to intersect BP produce at Q such that angle ABQ=ACQ PROVE THAT ANGLE AQC=90+1÷2anleBAC

Answer 1

I hope you are satisfied my answer...

Answer 2

Answer:

(Proof)

Step-by-step explanation:

See attached figure

Since ABD and ACD are angles in the same segment of chord AD, points A, B, C and D have to lie on the circle, i.e., they have to be concyclic.

AB = AC

∠ABC = ∠ACB

In triangle ABC,

∠ABC + ∠BAC + ∠ACB = 180°

∴∠BAC + 2∠ACB = 180°

2∠ACB = 180 - ∠BAC

∠ACB = $\frac{180 - BAC}{2}$

∠ACB = 90 - $\frac{1}{2}$BAC - - - -(i)

∠ACB = ∠ADB (ii) ∵Angles in the same segment

And ∠BAC = ∠BDC (iii) ∵Angles in the same segment

Adding (ii) and (iii)

∠ACB + ∠BAC = ∠ADB + ∠ BDC

∠ACB + ∠BAC = ∠ADC

∠ACB = ∠ADC - ∠BAC (iv)

Plugging value of ∠ACB from (i) in (iv),

∠ADC - ∠BAC = 90 - $\frac{1}{2}$BAC

∠ADC = 90 - $\frac{1}{2}$BAC + 1∠BAC

∠ADC = 90 + $\frac{1}{2}$BAC

==================

Hence proved.